Elements of geometry and trigonometry . to each other as their bases ;hence ADE : DEC : : AE : EC. But the triangles BDE. DEC, are equivalent ; and therefore,we have (Book II. Prop. IV. Cor.) AD : DB : : AE : EC. Cor. 1. Hence, by composition, we have AD + DB : AD : :AE + EC : AE, or AB : AD : : AC : AE ; and also AB :BD : : AC : CE. Cor. 2. If between two straight lines AB, CD, any numberof parallels AC, EF, GH, BD, &:c. be drawn, those straightlines w^ill be cut proportionally, and we shall have AE : CF ; :EG : FH : GB : HD. For, let O be the point where AB andCD meet. In the triangle OEF, t
Elements of geometry and trigonometry . to each other as their bases ;hence ADE : DEC : : AE : EC. But the triangles BDE. DEC, are equivalent ; and therefore,we have (Book II. Prop. IV. Cor.) AD : DB : : AE : EC. Cor. 1. Hence, by composition, we have AD + DB : AD : :AE + EC : AE, or AB : AD : : AC : AE ; and also AB :BD : : AC : CE. Cor. 2. If between two straight lines AB, CD, any numberof parallels AC, EF, GH, BD, &:c. be drawn, those straightlines w^ill be cut proportionally, and we shall have AE : CF ; :EG : FH : GB : HD. For, let O be the point where AB andCD meet. In the triangle OEF, the lineAC being drawn parallel to the base EF,we shall have OE : AE : : OF : CF, orOE : OF : ; AE : CF. In the triangleOGII, we shall likewise have OE : EG: : OF : FII,orOE : OF : : EG : by reason of the common ratio OE :OF, those two proportions give AE : CF: : EG : FH. It may be proved in thesame manner, that EG : FH : : GB : HD, and so on ; hencethe lines AB, CD, are cut proportionallv by the parallels AC,EF, GH, & BOOK IV. 83 PROPOSITION XVI. THEOREM. Conversely, if two sides of a triangle are cut proportiunally hy astraight line, this straight line will be parallel to the third side. In llie tiiaiigle ABC, let the line I)E be drawn, makingAD : DB : : AE : EC : then will DE be parallel to BC. For, if DE is not parallel to BC, draw DO paral-lel to it. Then, by the preceding theorem, we shallhave AD : DB : : AO : OC. But by hypothe-sis, we have AD : DB : : AE : EC : hence wemust have AO : OC : : AE : EC,orAO : AE: : OC : EC ; an impossible result, since AO, theone antecedent, is less than its consequent AE,and OC, the other antecedent, is greater than itsconsequent EC. Hence the parallel to BC, drawn from thepoint D, cannot ditler from DE ; hence DE is that parallel. Scholium. The same conclusion would be true, if the pro-portion AB : AD : : AC : AE were the proposed one. Forthis proportion would give AB—AD : AD : : AC—AE :.\E, or BD : AD : : CE : AE.
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